A bounded 6 variable multinomial inequality I am trying to show the following inequality:
$$2-DBC-AEC-ABF+ABCDEF\geq 0$$
where $A,B,C,D,E,F\in[0,1]$. It seems to be true from desmos and it passes a million random test cases on Mathematica. If it helps, we can also assume that $A>D,B>E,C>F$, but it doesn't seem necessary from the numerical tests.
My current approach is to stumble upon a nice factorization, but this has not been very fruitful. Are there other general methods for proving these sorts of non homogeneous multinomial inequalities?
 A: Start with $0< x,y,z< 1$ and note certainly $$\frac 1{1-x}+\frac 1{1-y}+\frac 1{1-z}>1,$$ or $$(1-y)(1-z)+(1-x)(1-z)+(1-x)(1-y)>(1-x)(1-y)(1-z).$$ Expanding out, this is $$3-2x-2y-2z+yz+xz+xy>1-x-y-z+xy+xz+yz-xyz,$$ or $$2+xyz>x+y+z.$$ Now we can put $x=DBC,$ $y=AEC$ and $z=ABF$ since by hypotheses, the right hand expressions are each between $0$ and $1.$ We get $$DBC+AEC+ABF<2+A^2B^2C^2DEF<2+ABCDEF,$$ where the last step follows because $ABC<1.$
A: $$
\begin{align}
2-DBC-AEC-ABF+ABCDEF &= 2 - \underbrace{F}_{\le 1} \cdot \underbrace{AB(1-CDE)}_{\ge 0}-DBC-AEC
\\ &\ge 2 - 1 \cdot AB(1-CDE) - DBC-AEC
\\ &= 2 - AB -DBC-AEC+ABCDE
\\ &= 2 - AB - E \cdot AC(1-BD) - DBC
\\ &\ge 2 - AB - 1 \cdot AC(1-BD) - DBC
\\ &= 2 - AB - AC - BCD + ACBD
\\ &= 2 - AB-AC-D \cdot BC(1-A)
\\ &\ge 2 - AB-AC-1 \cdot BC(1-A)
\\ &= 2 - AB - AC - BC + ABC
\\ &= 2 - \underbrace{\big(\; A\cdot\underbrace{(B+C)}_{\in\, [0,2]} + (1-A) \cdot \underbrace{BC}_{\in\,[0,1]} \;\big)}_{\text{weighted average}\; \in\,[BC, B+C]\,\subset\,[0,2]}
\\ &\ge 0
\end{align}
$$
The minimum value of $0$ is attained when all $\,\ge\,$ become $\,=\,$ iff $\,A=B=C=D=E=F=1$.
